;)
Hola karla!
Manda la segunda en otra pregunta y aclarame si es:
$$\begin{align}&r= \frac 1 {sen \theta-\cos \theta}\\&\\&o \\&\\&r=\frac 1 {sen \theta}-\cos \theta\end{align}$$
1.-
Las equivalencias son:
$$\begin{align}&r^2=x^2+y^2\\&\\&tan \theta= \frac y x\\&\\&r^2=sen(2 \theta)\\&\\&r^2= 2 sen \theta·\cos \theta\\&\\&trigonometría:\\&1+ tan^2 \theta=sec^2 \theta\\&\\&1+ (\frac y x)^2=sec^2 \theta\\&\frac{x^2+y^2}{x^2}= \frac 1 {\cos^2 \theta}\\&\\&\cos^2 \theta= \frac {x^2}{x^2+y^2}\\&\\&\cos \theta= \frac x {\sqrt {x^2+y^2}}\\&\\&sen^2 \theta=1-\cos^2 \theta=1-\frac {x^2}{x^2+y^2}=\frac{y^2}{x^2+y^2}\\&\\&sen \theta= \frac y { \sqrt {x^2+y^2}}\\&\\&r^2=sen(2 \theta)\\&\\&r^2=2 sen \theta \cos \theta\\&\\&x^2+y^2=2 \frac y { \sqrt {x^2+y^2}}\frac x { \sqrt {x^2+y^2}}\\&\\&\\&(x^2+y^2)^2=2xy\end{align}$$
Saludos
;)
:)